Exercise 33, § 1

Let E be a uniformly convex Banach space (V, p. 66, Exercise 31). There then exists a number $\theta$ such that $\frac{3}{4} \leqslant \theta < 1$ and such that the relation $\|x - y\| \geqslant \frac{1}{2} \sup(\|x\|, \|y\|)$ in E entails $\| \frac{1}{2}(x + y) \| \leqslant \theta \sup(\|x\|, \|y\|)$. a) Let $(x_n)$ be a sequence of points of E such that $\|x_n\| \leqslant M$ and which tends to 0 for $\sigma(E, E')$. Prove now that if for an index $p$, one has $\|x_p\| \geqslant \frac{1}{2}M$, there exists $q > p$ such that $\|x_p - x_q\| > \frac{1}{2}M$, and consequently $\| \frac{1}{2}(x_p + x_q) \| \leqslant \theta M$ (reason by contradiction, observing that for $x' \in E'$ such that $\|x'\| = 1$, one has $\langle x_p, x' \rangle = \lim_{n \to \infty} \langle x_p - x_n, x' \rangle$). Deduce that there exists a strictly increasing mapping $\ell$ of $\mathbf{N}$ into itself such that one has $\| \frac{1}{2}(x_{\ell(2n)} + x_{\ell(2n+1)}) \| \leqslant M\theta$, so that if $x_n^{(1)} = \frac{1}{2}(x_{\ell(2n)} + x_{\ell(2n+1)})$, the sequence $(x_n^{(1)})$ tends to 0 for $\sigma(E, E')$, and that $\|x_n^{(1)}\| \leqslant M\theta$ for every $n$. b) Prove now that there exists a sequence $(x_{n_k})$ extracted from $(x_n)$ such that, if one sets $y(k) = x_{n_k}$, it has the following property: for every integer $p > 1$, every integer $q < p$ and every integer $i$ such that $1 \leqslant i \leqslant 2^{p-q}$,

(Iterate the procedure of a) by forming from a sequence $(x_n^{(k)})$ a sequence $(x_n^{(k+1)})$ in the same way as $(x_n^{(1)})$ is formed from $(x_n)$; then use a suitable “diagonal procedure”.) c) Let $r$ and $q$ be two integers $> 1$. Deduce from b) that if $r2^q \leqslant k \leqslant (r+1)2^q$, one has (decompose the sum on the left into several parts, letting $h$ vary from 1 to $2^q$, then from $(j-1)2^q + 1$ to $j2^q$ for $2 \leqslant j \leqslant r$, then from $r2^q + 1$ to $k$). d) Prove now that for every bounded sequence $(x_n)$ in E, there exists an extracted sequence $(x_{n_k})$ such that the sequence of means $(x_{n_1} + \cdots + x_{n_k})/k$ converges for the initial topology of E (Banach-Saks-Kakutani theorem). (Using the fact that E is reflexive, reduce to the case where the sequence $(x_n)$ converges to 0 for $\sigma(E, E')$, and use c) for $q$ and $r$ sufficiently large.)